CHAPTER 4:
PROBABILITY
1
EXPERIMENT, OUTCOMES,
AND SAMPLE SPACE

Simple and Compound Events
2
EXPERIMENT, OUTCOMES,
AND SAMPLE SPACE
Definition
An experiment is a process that, when
performed, results in one and only one of
many observations. These observations are
called that outcomes of the experiment.
The collection of all outcomes for an
experiment is called a sample space.
3
Table 4.1
Experiment
Toss a coin once
Roll a die once
Toss a coin twice
Play lottery
Take a test
Select a student
Examples of Experiments, Outcomes, and
Sample Spaces
Outcomes
Head, Tail
1, 2, 3, 4, 5, 6
HH, HT, TH, TT
Win, Lose
Pass, Fail
Male, Female
Sample Space
S = {Head, Tail}
S = {1, 2, 3, 4, 5, 6}
S = {HH, HT, TH, TT}
S = {Win, Lose}
S = {Pass, Fail}
S = {Male, Female}
4
Example 4-1
Draw the Venn and tree diagrams for the
experiment of tossing a coin once.
5
Figure 4.1
(a) Venn Diagram and (b) tree diagram for
one toss of a coin.
Outcomes
Head
S
H
T
H
Tail
(a)
T
(b)
6
Example 4-2
Draw the Venn and tree diagrams for the
experiment of tossing a coin twice.
7
Figure 4.2 a
Venn diagram for two tosses of a coin.
S
HT
HH
TH
TT
(a)
8
Figure 4.2 b
First toss
Tree diagram for two tosses of coin.
Second
toss
H
Final
outcomes
HH
H
T
T
H
HT
TH
T
TT
(b)
9
Example 4-3
Suppose we randomly select two persons
from the members of a club and observe
whether the person selected each time is a
man or a woman. Write all the outcomes for
this experiment. Draw the Venn and tree
diagrams for this experiment.
10
Figure 4.3 a
Venn diagram for selecting two
persons.
S
MM
MW
WM
WW
(a)
11
Figure 4.3 b
First
selection
Tree diagram for selecting two
persons.
Second
selection
M
Final
outcomes
MM
M
W
W
M
W
MW
WM
WW
(b)
12
Simple and Compound Events
Definition
An event is a collection of one or more of
the outcomes of an experiment.
13
Simple and Compound Events
cont.
Definition
An event that includes one and only one of
the (final) outcomes for an experiment is
called a simple event and is denoted by
Ei.
14
Example 4-4
Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether
the person selected each time is a man or a
woman. Each of the final four outcomes (MM, MW,
WM, WW) for this experiment is a simple event.
These four events can be denoted by E1, E2, E3, and
E4, respectively. Thus,
E1 = (MM ), E2 = (MW ), E3 = (WM ), and
E4 = (WW )
15
Simple and Compound Events
Definition
A compound event is a collection of more
than one outcome for an experiment.
16
Example 4-5
Reconsider Example 4-3 on selecting two persons from the
members of a club and observing whether the person
selected each time is a man or a woman. Let A be the
event that at most one man is selected. Event A will occur
if either no man or one man is selected. Hence, the event
A is given by
A = {MW, WM, WW}
Because event A contains more than one outcome, it is a
compound event. The Venn diagram in Figure 4.4 gives a
graphic presentation of compound event A.
17
Figure 4.4
Venn diagram for event A.
S
A
MW
MM
WM
WW
18
Example 4-6
In a group of a people, some are in favor of genetic
engineering and others are against it. Two persons are
selected at random from this group and asked whether
they are in favor of or against genetic engineering. How
many distinct outcomes are possible? Draw a Venn
diagram and a tree diagram for this experiment. List all the
outcomes included in each of the following events and
mention whether they are simple or compound events.
(a) Both persons are in favor of the genetic engineering.
(b) At most one person is against genetic engineering.
(c) Exactly one person is in favor of genetic engineering.
19
Solution 4-6
Let






F = a person is in favor of genetic engineering
A = a person is against genetic engineering
FF = both persons are in favor of genetic engineering
FA = the first person is in favor and the second is
against
AF = the first is against and the second is in favor
AA = both persons are against genetic engineering
20
Figure 4.5 a
Venn diagram.
S
FF
FA
AF
AA
(a)
21
Figure 4.5 b
First
person
Tree diagram.
Second
person
F
Final
outcomes
FF
F
A
A
F
A
FA
AF
AA
(b)
22
Solution 4-6
Both persons are in favor of genetic
engineering = { FF }
It is a simple event.
b)
At most one person is against genetic
engineering = { FF, FA, AF }
It is a compound event.
c)
Exactly one person is in favor of genetic
engineering = { FA, AF }
It is a compound event.
a)
23
CALCULATING PROBABILITY


Two Properties of probability
Three Conceptual Approaches to
Probability



Classical Probability
Relative Frequency Concept of Probability
Subjective Probability
24
CALCULATING PROBABLITY
Definition
Probability is a numerical measure of the
likelihood that a specific event will occur.
25
Two Properties of Probability

First Property of Probability



0 ≤ P (Ei) ≤ 1
0 ≤ P (A) ≤ 1
Second Property of Probability

ΣP (Ei) = P (E1) + P (E2) + P (E3) + … = 1
26
Three Conceptual Approaches
to Probability
Classical Probability
Definition
Two or more outcomes (or events) that
have the same probability of occurrence
are said to be equally likely outcomes
(or events).
27
Classical Probability
Classical Probability Rule to Find Probability
1
P( Ei ) 
Total number of outcomes for the experiment
Number of outcomes favorable to A
P( A) 
Total number of outcomes for the experiment
28
Example 4-7
Find the probability of obtaining a head and
the probability of obtaining a tail for one
toss of a coin.
29
Solution 4-7
1
1
P(head ) 
  .50
Total number of outcomes 2
Similarly,
1
P( tail)   .50
2
30
Example 4-8
Find the probability of obtaining an even
number in one roll of a die.
31
Solution 4-8
P(head ) 
Number of outcomes included in A 3
  .50
Total number of outcomes
6
32
Example 4-9
In a group of 500 women, 80 have played
golf at lest once. Suppose one of these 500
women is randomly selected. What is the
probability that she has played golf at least
once?
33
Solution 4-9
80
P(selected woman has played golf at least once) 
 .16
500
34
Three Conceptual Approaches
to Probability cont.
Relative Concept of Probability
Using Relative Frequency as an Approximation
of Probability
If an experiment is repeated n times and an
event A is observed f times, then, according to
the relative frequency concept of probability:
f
P ( A) 
n
35
Example 4-10
Ten of the 500 randomly selected cars
manufactured at a certain auto factory are
found to be lemons. Assuming that the
lemons are manufactured randomly, what is
the probability that the next car
manufactured at this auto factory is a
lemon?
36
Solution 4-10
Let n denotes the total number of cars in the
sample and f the number of lemons in n. Then,
n = 500 and f = 10
Using the relative frequency concept of
probability, we obtain
P(next car is a lemon) 
f
10

 .02
n 500
37
Table 4.2
Frequency and Relative Frequency
Distributions for the Sample of Cars
Car
Good
Lemon
f
490
10
n = 500
Relative frequency
490/500 = .98
10/500 = .02
Sum = 1.00
38
Law of Large Numbers
Definition
Law of Large Numbers If an experiment
is repeated again and again, the probability
of an event obtained from the relative
frequency approaches the actual or
theoretical probability.
39
Three Conceptual Approaches
to Probability
Subjective Probability
Definition
Subjective probability is the probability
assigned to an event based on subjective
judgment, experience, information and
belief.
40
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if
the first step can result in m outcomes, the
second step in n outcomes, and the third in k
outcomes, then
Total outcomes for the experiment = m · n · k
41
Example 4-12
Suppose we toss a coin three times. This
experiment has three steps: the first toss, the
second toss and the third toss. Each step has two
outcomes: a head and a tail. Thus,
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are
HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT
42
Example 4-13
A prospective car buyer can choose between
a fixed and a variable interest rate and can
also choose a payment period of 36 months,
48 months, or 60 months. How many total
outcomes are possible?
43
Solution 4-13
Total outcomes = 2 x 3 = 6
44
Example 4-14
A National Football League team will play 16 games
during a regular season. Each game can result in
one of three outcomes: a win, a lose, or a tie. The
total possible outcomes for 16 games are calculated
as follows:
Total outcomes = 3·3·3·3·3·3·3·3·3·3·3·3 ·3·3·3·3
= 316 = 43,046,721
One of the 43,046,721 possible outcomes is all 16
wins.
45
MARGINAL AND CONDITIONAL
PROBABILITIES
Suppose all 100 employees of a company were
asked whether they are in favor of or against
paying high salaries to CEOs of U.S. companies.
Table 4.3 gives a two way classification of the
responses of these 100 employees.
46
Table 4.3
Male
Female
Two-Way Classification of Employee
Responses
In Favor
15
4
Against
45
36
47
MARGINAL AND CONDITIONAL
PROBABILITIES
Table 4.4
Male
Female
Total
Two-Way Classification of Employee Responses
with Totals
In Favor
15
4
19
Against
45
36
81
Total
60
40
100
48
MARGINAL AND CONDITIONAL
PROBABILITIES
Definition
Marginal probability is the probability of
a single event without consideration of any
other event. Marginal probability is also
called simple probability.
49
Table 4.5
Listing the Marginal Probabilities
In Favor Against Total
(A )
(B )
Male (M )
Female (F )
Total
15
4
19
45
36
81
60
40
100
P (M ) = 60/100 = .60
P (F ) = 40/100 = .40
P (A ) = 19/100 P (B ) = 81/100
= .19
= .81
50
MARGINAL AND CONDITIONAL
PROBABILITIES cont.
Read as “given”
P ( in favor | male)
The event whose
probability is to be
determined
This event has
already occurred
51
MARGINAL AND CONDITIONAL
PROBABILITIES cont.
Definition
Conditional probability is the probability that
an event will occur given that another has already
occurred. If A and B are two events, then the
conditional probability A given B is written as
P(A|B)
and read as “the probability of A given that B has
already occurred.”
52
Example 4-15
Compute the conditional probability
P ( in favor | male) for the data on 100
employees given in Table 4.4.
53
Solution 4-15
Male
In Favor
15
Males who are
in favor
Against
45
Total
60
Total number of
males
Number of males who are in favor 15
P(in favor | male) 

 .25
Total number of males
60
54
Figure 4.6
We are to find the
probability of this event
Tree Diagram.
This event has
already occurred
le
Ma
|
rs
vo
a
/60
F
15
Required probability
Ag a
le
Ma 00
/1
60
Fe
ma
40
/10 le
0
inst
45/6 | Male
0
ors
Fav
le
ma
e
F
|
Ag
4/4
0
ain
st |
Fem
ale
36/4
0
55
Example 4-16
For the data of Table 4.4, calculate the
conditional probability that a randomly
selected employee is a female given that
this employee is in favor of paying high
salaries to CEOs.
56
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
Number of females who are in favor
P(female | in favor) 
Total number of employees who are in favor
4

 .2105
19
57
Figure 4.7
This event has
already occurred
Tree diagram.
rs
vo
a
|F
le
a
/60
M
15
Fem
a
4/19
rs
vo
Fa 100
/
19
Ag
81 ainst
/10
0
We are to find the
probability of this event
le |
Fav
o
rs
Required probability
le
Ma
|
ai
Ag
45
nst
/81
Fem
ale
36/8
1
|A
gai
nst
58
MUTUALLY EXCLUSIVE
EVENTS
Definition
Events that cannot occur together are said
to be mutually exclusive events.
59
Example 4-17
Consider the following events for one roll of a
die:
A= an even number is observed= {2, 4, 6}
B= an odd number is observed= {1, 3, 5}
C= a number less than 5 is observed= {1, 2, 3, 4}
Are events A and B mutually exclusive? Are
events A and C mutually exclusive?
60
Solution 4-17
Figure 4.8
Mutually exclusive events A and B.
S
A
1
2
6
5
3
4
B
61
Solution 4-17
Figure 4.9
Mutually nonexclusive events A and C.
62
Example 4-18
Consider the following two events for a
randomly selected adult:
Y = this adult has shopped on the Internet at
least once
N = this adult has never shopped on the Internet
Are events Y and N mutually exclusive?
63
Solution 4-18
Figure 4.10
Mutually exclusive events Y and N.
S
Y
N
64
INDEPENDENT VERSUS
DEPENDENT EVENTS
Definition
Two events are said to be independent if the
occurrence of one does not affect the
probability of the occurrence of the other. In
other words, A and B are independent
events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
65
Example 4-19
Refer to the information on 100 employees
given in Table 4.4. Are events “female (F )”
and “in favor (A )” independent?
66
Solution 4-19
Events F and A will be independent if
P (F ) = P (F | A )
Otherwise they will be dependent.
From the information given in Table 4.4
P (F ) = 40/100 = .40
P (F | A ) = 4/19 = .2105
Because these two probabilities are not equal, the
two events are dependent.
67
Example 4-20
A box contains a total of 100 CDs that were
manufactured on two machines. Of them, 60 were
manufactured on Machine I. Of the total CDs, 15
are defective. Of the 60 CDs that were
manufactured on Machine I, 9 are defective.
Let D be the event that a randomly selected CD is
defective, and let A be the event that a randomly
selected CD was manufactured on Machine I. Are
events D and A independent?
68
Solution 4-20
From the given information,
P (D ) = 15/100 = .15
P (D | A ) = 9/60 = .15
Hence,
P (D ) = P (D | A )
Consequently, the two events are
independent.
69
Table 4.6
Two-Way Classification Table
Defective
(D )
Machine I (A )
Machine II (B )
Total
Good
(G )
Total
9
6
51
34
60
40
15
85
100
70
Two Important Observations

Two events are either mutually exclusive or
independent.



Mutually exclusive events are always
dependent.
Independent events are never mutually
exclusive.
Dependents events may or may not be
mutually exclusive.
71
COMPLEMENTARY EVENTS
Definition
The complement of event A, denoted by Ā
and is read as “A bar” or “A complement”, is
the event that includes all the outcomes for
an experiment that are not in A.
72
Figure 4.11
Venn diagram of two complementary
events.
S
A
A
73
Example 4-21
In a group of 2000 taxpayers, 400 have
been audited by the IRS at least once. If one
taxpayer is randomly selected from this
group, what are the two complementary
events for this experiment, and what are
their probabilities?
74
Solution

The complementary events for this experiment are



A = the selected taxpayer has been audited by
the IRS at least once
Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are:
P (A) = 400/2000 = .20
P (Ā) = 1600/2000 = .80
75
Figure 4.12
Venn diagram.
S
A
A
76
Example 4-22
In a group of 5000 adults, 3500 are in favor
of stricter gun control laws, 1200 are
against such laws, and 300 have no opinion.
One adult is randomly selected from this
group. Let A be the event that this adult is
in favor of stricter gun control laws. What is
the complementary event of A? What are
the probabilities of the two events?
77
Solution 4-22

The two complementary events are



A = the selected adult is in favor of stricter gun
control laws
Ā = the selected adult either is against such laws or
has no opinion
The probabilities of the complementary events
are:
P (A) = 3500/5000 = .70
P (Ā) = 1500/5000 = .30
78
Figure 4.13
Venn diagram.
S
A
A
79
INTERSECTION OF EVENTS AND
THE MULTIPLICATION RULE


Intersection of Events
Multiplication Rule
80
Intersection of Events
Definition
Let A and B be two events defined in a
sample space. The intersection of A and B
represents the collection of all outcomes that
are common to both A and B and is denoted
by
A and B
81
Figure 4.14
Intersection of events A and B.
B
A
A
and
B
Intersection of A and B
82
Multiplication Rule
Definition
The probability of the intersection of two
events is called their joint probability. It is
written as
P (A and B ) or P (A ∩ B )
83
INTERSECTION OF EVENTS AND
THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two
events A and B is
P (A and B ) = P (A )P (B |A )
84
Example 4-23
Table 4.7 gives the classification of all
employees of a company given by gender
and college degree.
85
Table 4.7
Male (M )
Female (F )
Total
Classification of Employees by Gender and
Education
College
Graduate
(G )
7
4
Not a College
Graduate
(N )
20
9
Total
27
13
11
29
40
86
Example 4-23
If one of these employees is selected at
random for membership on the employee
management committee, what is the
probability that this employee is a female
and a college graduate?
87
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F )
P (F ) = 13/40
P (G |F ) = 4/13
P (F and G ) = (13/40)(4/13) = .100
88
Figure 4.15
Intersection of events F and G.
Females
College graduates
4
Females and college graduates
89
Figure 4.16
Male / female
Tree diagram for joint probabilities.
Graduates / nongraduates
Final outcomes
P(M and G) = (27/40) (20/27) = .175
G|M
7/27
N|M
M
20/27
P(M and N) = (27/40) (20/27) = .500
24/40
F
13/40
P(F and G) = (13/40) (4/13) = .100
G|F
4/13
9/13
N|F
P(F and N) = (13/40) (9/13) = .225
90
Example 4-24
A box contains 20 DVDs, 4 of which are
defective. If 2 DVDs are selected at random
(without replacement) from this box, what is
the probability that both are defective?
91
Solution 4-24
Let us define the following events for this
experiment:
G1 = event that the first DVD selected is good
D1 = event that the first DVD selected is defective
G2 = event that the second DVD selected is good
D2 = event that the second DVD selected is defective
The probability to be calculated is
P (D1 and D2) = P (D1 )P (D2 |D1 )
P (D1) = 4/20
P (D2 |D1) = 3/19
P (D1 and D2) = (4/20)(3/19) = .0316
92
Figure 4.17
First selection
Selecting two DVDs.
Second selection
G2 | G1
Final outcomes
P(G1 and G2) = (16/20) (15/19) = .6316
15/19
D2 | G1
G1
4/19
16/20
D1
4/20
G2 | D1
P(G1 and D2) = (16/20) (4/19) = .1684
P(D1 and G2) = (4/20) (16/19) = .1684
16/19
D2 | D1
3/19
P(D1 and D2) = (4/20) (3/19) = .0316
93
Multiplication Rule cont.
Calculating Conditional Probability
If A and B are two events, then,
P( A and B)
P( A and B)
P( B | A) 
and P( A | B) 
P( A)
P( B)
given that P (A ) ≠ 0 and P (B ) ≠ 0.
94
Example 4-25
The probability that a randomly selected
student from a college is a senior is .20, and
the joint probability that the student is a
computer science major and a senior is .03.
Find the conditional probability that a
student selected at random is a computer
science major given that he/she is a senior.
95
Solution 4-25

Let us define the following two events:


A = the student selected is a senior
B = the student selected is a computer science
major


From the given information,
P (A) = .20 and P (A and B) = .03
Hence,
P (B | A ) = .03/.20 = .15
96
Multiplication Rule for
Independent Events
Multiplication Rule to Calculate the
Probability of Independent Events
The probability of the intersection of two
independent events A and B is
P (A and B ) = P (A )P (B )
97
Example 4-26
An office building has two fire detectors. The
probability is .02 that any fire detector of
this type will fail to go off during a fire. Find
the probability that both of these fire
detectors will fail to go off in case of a fire.
98
Solution 4-26
Let
A = the first fire detector fails to go off during a fire
B = the second fire detector fails to go off during a
fire
Then, the joint probability of A and B is
P (A and B ) = P (A) P (B ) = (.02)(.02) = .0004
99
Example 4-27
The probability that a patient is allergic to
penicillin is .20. Suppose this drug is
administered to three patients.
a) Find the probability that all three of them are
allergic to it.
b) Find the probability that at least one of the
them is not allergic to it.
100
Solution
a)
Let A, B, and C denote the events the first,
second and third patients, respectively, are
allergic to penicillin. Hence,
P (A and B and C ) = P (A ) P (B ) P (C )
= (.20) (.20) (.20) = .008
101
Solution
b)
Let us define the following events:
G = all three patients are allergic
H = at least one patient is not allergic



P (G ) = P (A and B and C ) = .008
Therefore, using the complementary event
rule, we obtain
P (H ) = 1 – P (G )
= 1 - .008 = .992
102
Figure 4.18
First patient
Tree diagram for joint probabilities.
Second patient
Third patient
C
Final outcomes
P(ABC) = .008
.20
B
.20
C
.80
P(ABC) = .032
B
A
.80
.20
A
.80
C
.20
P(ABC) = .032
C
.80
P(ABC) = .128
C
P(ABC) = .032
.20
B
.20
C
.80
P(ABC) = .128
B
.80
C
.20
P(ABC) = .128
C
.80
P(ABC) = .512
103
Multiplication Rule for
Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually
exclusive events is always zero. If A and B
are two mutually exclusive events, then
P (A and B ) = 0
104
Example 4-28
Consider the following two events for an
application filed by a person to obtain a car
loan:
A = event that the loan application is approved
R = event that the loan application is rejected
What is the joint probability of A and R?
105
Solution 4-28
The two events A and R are mutually
exclusive. Either the loan application will be
approved or it will be rejected. Hence,
P (A and R ) = 0
106
UNION OF EVENTS AND THE
ADDITION RULE
Definition
Let A and B be two events defined in a
sample space. The union of events A and B
is the collection of all outcomes that belong
to either A or B or to both A and B and is
denoted by
A or B
107
Example 4-29
A senior citizen center has 300 members.
Of them, 140 are male, 210 take at least
one medicine on a permanent basis, and 95
are male and take at least one medicine on
a permanent basis. Describe the union of
the events “male” and “take at least one
medicine on a permanent basis.”
108
Solution 4-29

Let us define the following events:
M = a senior citizen is a male
F = a senior citizen is a female
A = a senior citizen takes at least one medicine
B = a senior citizen does not take any medicine

The union of the events “male” and “take at least
one medicine” includes those senior citizens who
are either male or take at least one medicine or
both. The number of such senior citizen is
140 + 210 – 95 = 255
109
Table 4.8
M
F
Total
A
B
95
115
210
45
45
90
Total
140
160
300
Counted twice
110
Figure 4.19
M
Union of events M and A.
A
Shaded area gives the union of events M and A, and
includes 255 senior citizen
111
Multiplication Rule for
Independent Events
Addition Rule
Addition Rule to Find the Probability of Union
of Events
The portability of the union of two events A
and B is
P (A or B ) = P (A ) + P (B) – P (A and B )
112
Example 4-30
A university president has proposed that all
students must take a course in ethics as a
requirement for graduation. Three hundred faculty
members and students from this university were
asked about their opinion on this issue. Table 4.9
gives a two-way classification of the responses of
these faculty members and students.
Find the probability that one person selected at
random from these 300 persons is a faculty
member or is in favor of this proposal.
113
Table 4.9
Two-Way Classification of Responses
Favor
Oppose
Neutral
Total
Faculty
Student
45
90
15
110
10
30
70
230
Total
135
125
40
300
114
Solution 4-30
Let us define the following events:
A = the person selected is a faculty member
B = the person selected is in favor of the proposal
From the information in the Table 4.9,
P (A ) = 70/300 = .2333
P (B ) = 135/300 = .4500
P (A and B) = P (A) P (B | A ) = (70/300)(45/70) = .1500
Using the addition rule, we have
P (A or B ) = P (A ) + P (B ) – P (A and B )
= .2333 + .4500 – .1500 = .5333
115
Example 4-31
A total of 2500 persons, 1400 are female,
600 are vegetarian, and 400 are female and
vegetarian. What is the probability that a
randomly selected person from this group is
a male or vegetarian?
116
Solution 4-31
Let us define the following events:
F = the randomly selected person is a female
M = the randomly selected person is a male
V = the randomly selected person is a vegetarian
N = the randomly selected person is a non-vegetarian.
P( M or V )  P( M )  P(V )  P( M and V )
1100 600
200



2500 2500 2500
 .44  .24  .08  .60
117
Table 4.10
Two-Way Classification Table
Vegetarian (V) Nonvegetarian (N)
Total
Female (F)
Male (M)
400
200
1000
900
1400
1100
Total
600
1900
2500
118
Addition Rule for Mutually
Exclusive Events
Addition Rule to Find the Probability of the
Union of Mutually Exclusive Events
The probability of the union of two
mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
119
Example 4-32
A university president has proposed that all
students must take a course in ethics as a
requirement for graduation. Three hundred
faculty members and students from this university
were asked about their opinion on this issue. The
following table, reproduced from Table 4.9 in
Example 4-30, gives a two-way classification of
the responses of these faculty members and
students.
120
Table 4.9
Two-Way Classification of Responses
Favor
Oppose
Neutral
Total
Faculty
Student
45
90
15
110
10
30
70
230
Total
135
125
40
300
121
Example 4-32
What is the probability that a randomly
selected person from these 300 faculty
members and students is in favor of the
proposal or is neutral?
122
Figure 4.20
F
Venn diagram of mutually exclusive
events.
N
123
Solution 4-32
Let us define the following events:
F = the person selected is in favor of the proposal
N = the person selected is neutral
From the given information,
P (F ) = 135/300 = .4500
P (N ) = 40/300 = .1333
Hence,
P (F or N ) = P (F ) + P (N ) = .4500 + .1333 = .5833
124
Example 4-33
Consider the experiment of rolling a die
twice. Find the probability that the sum of
the numbers obtained on two rolls is 5, 7,
or 10.
125
Table 4.11
Two Rolls of a Die
Second Roll of the Die
First
Roll of
the Die
1
2
3
4
5
6
1
(1,1)
(1,2)
(1,3)
(1,4)
(1,5) (1,6)
2
(2,1)
(2,2)
(2,3)
(2,4)
(2,5) (2,6)
3
(3,1)
(3,2)
(3,3)
(3,4)
(3,5) (3,6)
4
(4,1)
(4,2)
(4,3)
(4,4)
(4,5) (4,6)
5
(5,1)
(5,2)
(5,3)
(5,4)
(5,5) (5,6)
6
(6,1)
(6,2)
(6,3)
(6,4)
(6,5) (6,6)
126
Solution 4-33
P (sum is 5 or 7 or 10)
= P (sum is 5) + P (sum is 7) + P (sum is 10)
= 4/36 + 6/36 + 3/36
= 13/36
= .3611
127
Example 4-34
The probability that a person is in favor of
genetic engineering is .55 and that a
person is against it is .45. Two persons are
randomly selected, and it is observed
whether they favor or oppose genetic
engineering.
a)
b)
Draw a tree diagram for this experiment
Find the probability that at least one of the
two persons favors genetic engineering.
128
Solution 4-34
a) Let
F = a person is in favor of genetic engineering
A = a person is against genetic engineering
The tree diagram in Figure 4.21 shows these four
outcomes and their probabilities.
129
Figure 4.21
First person
Tree diagram.
Final outcomes and their
probabilities
Second person
P(FF) = (.55) (.55) = .3025
F
.55
A
F
.45
.55
P(FA) = (.55) (.45) = .2475
.55
A
.45
P(AF) = (.45) (.55) = .2475
F
.55
A
.45
P(AA) = (.45) (.45) = .2025
130
Solution
b)
P ( at least one person favors)
= P (FF or FA or AF )
= P (FF ) + P (FA ) + P (AF )
= .3025 + .2475 + .2475
= .7975
131